Integrating a squared function (1 Viewer)

[bemer]

What is the integral of (4x+3)squared?

Should i expand then integrate

Or just integrate straight away

I get two different answer

Thanks for All Help

 

[kwabon]

[4x+3]^2
[4x+3]^3 / 4*3 + c


16x^2 + 24x + 9
16/3 x^3 + 24/2 x^2 + 9x + c



expand it out and hopefully, it will be the exact same thing.

now i see whats wrong, if you use the first method you get a 27/12 right? its a constant dw worry about it, that is why there is a +c in the second, if it was a definite integral, they would add up to be the exact same thing.

good luck

 

[bemer]

lol its different

 

[addikaye03]

What is the integral of (4x+3)squared?

Should i expand then integrate

Or just integrate straight away

I get two different answer

Thanks for All Help

you just include the coefficient of x in the denominator

int (4x+3)^2 dx

[(4x+3)^3]/ 4.3 + C

(1/12)(4x+3)^3 + C

 

[kwabon]

lol its different

edited my post.

 

[samaccount123]

Narh Bemer, its the same..its just the constant is found out if you use the other method..both are correct unless they want you to graph it - then you need the constant..or easier to just substitute values and equate to find C

 

[Trebla]

Keep in mind this reverse chain rule formula is only true if the inner function is linear. In other words if the derivative you divide is not constant then the formula does not hold.

 

[samaccount123]

Sup,

...lol, this was a long time since I done integration since 05.. anywayz.

I use another way to integrate it because I learnt this from Maths in focus book 2:

Since x in the function is to the power of 1 (invisible 1) after the coefficient of 4, you can use what addikaye has said (Which is only what 2 unit maths people should do - 3 unit maths people know another way of integrating this I think..thats what I remember)..

Firstly yeah, you gotta know what type of curve the function is and what type of integral you think its showing..either a trig integral or a exponential function..this one is basic.

Then check if its a function of a function or a chained function, this one is raised to the power of more than 1 and hence occurs more than once (i.e. to the power of 2 which happens twice)..so its a function of a function...forgot why but you gotta know what type of curve it is in general, lol.

Hence..

What is the integral of (4x+3) squared?

Q1: (4x+3)^2


let u = (4x+3)

Therefore

Q1 = u^2

So Integral of u^2:

= u^3 / 3 + C

= 1/3 u^3 + C (1)

Therefore sub. (1) into Q1: (4x+3) ^ 3 / 3 + C

= 1/3 (4x + 3) ^ 3 + C

However what addikaye told you now applies, this is an important rule that only 2 unit students should use - 3 units can use 2 methods, one of which we dont know how to do:

So, yes..just multiply the co-efficient of x in the question to the denominator ONLY to the integrated answer (Which is the 2 unit RULE that MUST be remembered)

= 1/3 (4x + 3) ^ 3 . 1/4 + C

= 1/ 3.4 (4 x + 3) ^ 3 + C

= 1/12 (4x + 3) ^ 3 + C

MY way of doing this part however is just remembering something I saw in the maths in focus book, instead of remembering the words about the coefficient - just remember this rule/ formulae.

Area under the curve function = The Integral of [F(x)]^n X 1/F'(x) + C

So simply integrate the function..which is:

1/3 (4x + 3) ^ 3 + C (as above)

THEN:

1/ F'(x) = 1/ (4)

THEREFORE:

1/3 (4x+3) ^ 3 X 1/4

= 1/12 (4x+3) ^ 3 + C

...lol

 

[samaccount123]

Keep in mind this reverse chain rule formula is only true if the inner function is linear. In other words if the derivative you divide is not constant then the formula does not hold.

Since x in the function is to the power of 1 (invisible 1) after the coefficient of 4, you can use what addikaye has said (Which is only what 2 unit maths people should do - 3 unit maths people know another way of integrating this I think..thats what I remember)..

Yep same thing , but you done a better technical work explaining it than me

 

[bemer]

Cheers everyone
I get it now the answer is
(4x+3)CUBED over 12 +c
is the correct answer

Much appreciated.

 

[samaccount123]

its not only the coefficient of x you have to include in the denominator, which is what addikaye has said previously..because there could be any other letter in there including the x.

e.g. (ax + by + c)..therefore it really is the inverse of the differential of the funtion multiplied by the integral of the question is what you really should do. But re-check that in your maths textbook to make sure.

Look like I've explained it the most, hah..St. JPHS 05 Maths FTW, yeah thats my HS..lol

 

[samaccount123]

Somethings wrong with the site today, things aren't clear

 

[Trebla]

its not only the coefficient of x you have to include in the denominator, which is what addikaye has said previously..because there could be any other letter in there including the x.

e.g. (ax + by + c)..therefore it really is the inverse of the differential of the funtion multiplied by the integral of the question is what you really should do. But re-check that in your maths textbook to make sure.

Look like I've explained it the most, hah..St. JPHS 05 Maths FTW, yeah thats my HS..lol

If you are integrating with respect to x, the co-efficient of y is pretty much irrelevant, unless y = f(x) i.e. y is a function of x in which case you can substitute y for an expression in terms of x.